Problem: The lifespans of bears in a particular zoo are normally distributed. The average bear lives $39.2$ years; the standard deviation is $7.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living between $54$ and $61.4$ years.
Explanation: $39.2$ $31.8$ $46.6$ $24.4$ $54$ $17$ $61.4$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $39.2$ years. We know the standard deviation is $7.4$ years, so one standard deviation below the mean is $31.8$ years and one standard deviation above the mean is $46.6$ years. Two standard deviations below the mean is $24.4$ years and two standard deviations above the mean is $54$ years. Three standard deviations below the mean is $17$ years and three standard deviations above the mean is $61.4$ years. We are interested in the probability of a bear living between $54$ and $61.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the bears will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the bears will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of bears between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular bear living between $54$ and $61.4$ years is $\color{orange}{2.35\%}$.